Question

In problems 1-20 find either F(s) or f(t), as indicated.

16.${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{2s+5}{s^2+6s+34}\right\}$

Short answer

The solution of the given Inverse Laplace transform is $e^{-3t}\left(2\cos \left(5t\right)-\frac{1}{5}\sin \left(5t\right)\right)$

Step-by-step solution

Definition of Inverse transforms

If $\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$represents the Laplace transform of a function$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, that is, $\mathit{L}\left\{f\left(t\right)\right\}\mathbf{=}\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$,we then say$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is the inverse Laplace transform of$\mathit{F}\mathbf{\left(}\mathit{s}\mathbf{\right)}$and write $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{F\left(s\right)\right\}$.

Inverse Laplace transform of given function

$L^{-1}\left\{\frac{2s+5}{s^2+6s+34}\right\}$

First,we need to see whether we can factor the denominator.

If we apply the quadratic formula, $s^2+6s+34=0$in,the equation has no real solution.

So the denominator cannot be simplified into a product of linear factors.

Now, to complete the square,

role="math" localid="1664223274308" $$\begin{gathered} s^2+6s+34=s^2+2.3.s+34 \\ \begin{array}{l}=s^2+2.3.s+9+25\\ ={\left(s+3\right)}^2+5^2\end{array} \end{gathered}$$

Now, rewrite the given function of in a more suitable way,

role="math" localid="1664223313196" $$\begin{gathered} \frac{2s+5}{s^2+6s+34}=\frac{2s+5}{{\left(s+3\right)}^2+5^2} \\ \begin{array}{c}=\frac{2s+6-1}{{\left(s+3\right)}^2+5^2}\\ =2\frac{s+3}{{\left(s+3\right)}^2+5^2}-\frac{1}{{\left(s+3\right)}^2+5^2}\end{array} \end{gathered}$$

Using the linearity of the inverse Laplace transform 

Now, using the linearity of the inverse Laplace transform, the first shift theorem and the

known Laplace transforms of the sine and cosine we get,

$\begin{array}{rcl}{L}^{-1}\left\{\frac{2s+5}{s^2+6s+34}\right\}& =& 2L^{-1}\left\{\frac{s+3}{{\left(s+3\right)}^2+5^2}\right\}-L^{-1}\left\{\frac{1}{{\left(s+3\right)}^2+5^2}\right\}\\ & =& 2e^{-3t}{L}^{-1}\left\{\frac{s}{{\left(s\right)}^2+5^2}\right\}-\frac{1}{5}{e}^{-3t}{L}^{-1}\left\{\frac{5}{{\left(s\right)}^2+5^2}\right\}\\ & =& 2e^{-3t}\cos \left(5t\right)-\frac{e^{-3t}}{5}\sin \left(5t\right)\\ & =& e^{-3t}\left(2\cos \left(5t\right)-\frac{1}{5}\sin \left(5t\right)\right)\\ & & \end{array}$

Thus, $L^{-1}\left\{\frac{2s+5}{s^2+6s+34}\right\}=e^{-3t}\left(2\cos \left(5t\right)-\frac{1}{5}\sin \left(5t\right)\right)$

Therefore, the solution of the given Inverse Laplace transform $L^{-1}\left\{\frac{2s+5}{s^2+6s+34}\right\}$is $e^{-3t}\left(2\cos \left(5t\right)-\frac{1}{5}\sin \left(5t\right)\right)$.